A model rocket is launched into the air. Its height, h(t), in metres after t seconds is h(t) = -5t2 + 40t +2?

a)
h(t) = -5t² + 40t + 2

When t = 2:
h(2) = -5(2)² + 40(2) + 2 = 62

Height of the rocket after 2 s = 64 m

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b)
When h(t) = 0:
-5t² + 40t + 2 = 0
t = {-40 ± √[40² - 4(-5)(2)]} / [2(-5)]
t = 8 or t = -0.05 (rejected)

Time taken for the rocket to hit the ground = 8 s

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c)
When h(t) = 77
-5t² + 40t + 2 = 77
-5t² + 40t - 75 = 0
t² - 8t + 15 = 0
(t - 3)(t - 5) = 0
t = 3 or t = 5

Time for the rocket at the height of 77 m = 3 s or 5 s

(a) h = -5 * 4 + 40 * 2 + 2 = 82 - 20 = 62 metres
(b) -5t^2 + 40t + 2 = 0
==> t = (-40 +/- sqrt(1600 + 40))/(-10)
==> t = 8.05 seconds
(c) 77 = -5t^2 + 40t + 2
==> 5t^2 - 40t + 75 = 0
==> t^2 -8t + 15 = 0
==> (t -3)(t -5) = 0
==> t = 3 or 5 seconds.

h(t) = -5t2 + 40t +2
I use the following equation for this type of problem.
h(t) = hi + vi * t – ½ * a * t^2

In this problem, the initial height is 2 meters. The initial velocity is 40 m/s. And the acceleration is -10 m/s^2. I will use this equation to check your answers.

h(f) = 2 + 40 * t – 5 * t^2
Let t equal 2 seconds
h(f) = 2 + 40 * 2 – 5 * 2^2 = 62 meters
For problem, the rock will hit the ground when its final height is 0 meters
0 = 2 + 40 * t – 5 * t^2
5 * t^2 – 40 * t – 2 = 0
t = [40 ± √(1600 – 4 * 5 * -2)] ÷ 10
t = [40 ± √1640] ÷ 10
t = [40 + √1640] ÷ 10
The time is approximately 8 seconds. The other answer is a negative time.

c.) When is the rocket at the height of 77m?
The rocket is at the height of 77m at 3 s and 5 s

For this problem, let h(f) be 77 meters. Use the same equation to determine the times.

77 = 2 + 40 * t – 5 * t^2
5 * t^2 – 40 * t + 75 = 0
Divide both sides of this equation by 5.

t^2 – 8 * t + 15 = 0
t = [8 ± √(64 – 4 * 1 * 15)] ÷ 2]
t = [8 ± 2)] ÷ 2
t1 = 5 seconds
t2 = 3 seconds
All of the answers are correct.